Let us start with observing that, by virtue of hypothesis (ii),
the bifunction \(d_C\circ f\)  is continuous on \(\mathbb {X}\times \mathbb {X}\) .
By the compactness of \(K\) ,
this implies that, for every \(x\in K\) , the value \(\nu (x)=\sup _{z\in K}{\rm dist}\left(f(x,z),C\right)\)  is attained at some \(z\in K\) ,
so that \(\overline{K}_x\ne \varnothing \)  for every \(x\in K\) .
Furthermore, the continuity of \(d_C\circ f\)  entails that
each function \(x\mapsto {\rm dist}\left(f(x,z),C\right)\)  is continuous on \(\mathbb {X}\) 
and each function \(z\mapsto {\rm dist}\left(f(x,z),C\right)\)  is, in particular,
u.s.c. on \(\mathbb {X}\) . Since by virtue of hypothesis (iii) each
function \(x\mapsto {\rm dist}\left(f(x,z),C\right)\)  is convex on \(\mathbb {X}\) 
(remember Lemma REF ), these facts enable one
to apply the well-known max rule for the subdifferential of
a supremum of convex functions (see, for instance,
[1]}), which gives
\( \partial \nu (x) &=& \partial \left(\sup _{z\in K}{\rm dist}\left(f(\cdot ,z),C\right)\right)(x) \nonumber \\&=& \overline{\rm conv}{\,}^*\, \left(\bigcup _{z\in \overline{K}_x}\partial {\rm dist}\left(f(\cdot ,z),C\right)(x)\right).\) 
Since each function \(x\mapsto (x,z)\)  is,
in particular, strictly differentiable and its derivative
is surjective according to hypothesis (iv), then it is possible
to employ the formula in [2]},
which rules the subdifferential under composition with
smooth mappings. In doing so, by recalling that the Mordukhovich
(a.k.a. basic or limiting) subdifferential
coincides here with the subdifferential in the sense of convex
analysis by the convexity of the involved functions, one finds
\(\partial {\rm dist}\left(f(\cdot ,z),C\right)(x)\subseteq {\rm D}f(\cdot ,z)(x)^*\partial d_C(f(x,z)).\) 
Now, observe that if \(x\in \mathbb {X}\backslash {\mathcal {S}}{\mathcal {E}}\)  and \(\nu (x)>0\) ,
then for every \(z\in \overline{K}_x\)  it must be \(f(x,z)\notin C\) .
Consequently, as also \((\mathbb {Y},\Vert \cdot \Vert )\)  is reflexive,
\(\Pi (f(x,z);C)\ne \varnothing \)  and therefore
\(\partial d_C(f(x,z))\subseteq \bigcap _{w\in \Pi (f(x,z);C)}{\rm N}(w;C)\cap {\mathbb {S}}^*\subseteq {C}^{{}^\ominus }\cap {\mathbb {S}}^*.\) 
If \(x\in \mathbb {X}\backslash {\mathcal {S}}{\mathcal {E}}\)  but \(\nu (x)=0\) , then it happens that
\(f(x,z)\in C\)  also for \(z\in \overline{K}_x\) , so one can only
say
\(\partial d_C(f(x,z))\subseteq {C}^{{}^\ominus }\cap {\mathbb {B}}^*.\) 
Thus, from inclusion \((\ref {in:wclcomaxform})\)  one obtains in any case
\(\partial \nu (x)\subseteq \overline{\rm conv}{\,}^*\, \left(\bigcup _{z\in \overline{K}_x}{\rm D}f(\cdot ,z)(x)^*(B^*_C(x))\right),\quad \forall x\in [\nu _{+K}>0].\) 
In the light of the last inclusion, it is clear that hypothesis (v)
implies the validity of condition (iv) in Theorem REF .
It remains to notice that, for any fixed \(x\in \mathbb {X}\) , by continuity of the
function \(z\mapsto {\rm dist}\left(f(x,z),C\right)\)  and by compactness of \(K\) ,
also the set \(f(x,K)\)  is compact and hence \(C\) -bounded, so
hypothesis (ii) of Theorem REF  is fulfilled.
As for hypothesis (iii) of Theorem REF , it comes
true as an easy consequence of the current hypothesis (ii).
Thus, the thesis can be achieved by applying Theorem REF .
